Problem: Let $f(x, y, z) = y^2\sin(x) +xz$ Suppose $\vec{a} = (0, 2, 1)$ and $\vec{v} = \left( 1, 0, -2 \right)$. Find the directional derivative of $f(x, y, z)$ at $\vec{a}$ in the direction of $\vec{v}$. Do not normalize the direction vector for your calculation. $\dfrac{\partial f}{\partial v} = $
The directional derivative of a function $f$ at $\vec{a}$ in the direction of $\vec{v}$ equals $\left( \nabla f (\vec{a}) \right) \cdot \vec{v}$. Let's find the gradient of $f$. $\begin{aligned} \nabla f &= \left( \dfrac{\partial f}{\partial x}, \dfrac{\partial f}{\partial y}, \dfrac{\partial f}{\partial z} \right) \\ \\ &= \left( y^2\cos(x) + z, 2y\sin(x), x \right) \end{aligned}$ Plugging in $(0, 2, 1)$, we can find $\nabla f$ at $\vec{a}$. $\nabla f (\vec{a}) = (5, 0, 0)$ Therefore: $\begin{aligned} \left( \nabla f (\vec{a}) \right) \cdot \vec{v} &= (5, 0, 0) \cdot (1, 0, -2) \\ \\ &= 5 \end{aligned}$ In conclusion, the directional derivative of $f$ at $\vec{a}$ in the direction of $\vec{v}$ equals $5$.